Question

the measure of angle a is 15°, and the length of side bc is 8. what are the lengths of the other two sides, rounded to the nearest tenth? ac = ab =

Explanation:

Step1: Use trigonometric ratios

In right - triangle \(ABC\) (\(\angle C = 90^{\circ}\)), \(\sin A=\frac{BC}{AB}\) and \(\tan A=\frac{BC}{AC}\).
Given \(\angle A = 15^{\circ}\) and \(BC = 8\).

Step2: Find \(AB\)

Since \(\sin A=\frac{BC}{AB}\), then \(AB=\frac{BC}{\sin A}\).
Substitute \(A = 15^{\circ}\) and \(BC = 8\) into the formula: \(AB=\frac{8}{\sin15^{\circ}}\).
We know that \(\sin15^{\circ}=\sin(45^{\circ}- 30^{\circ})=\sin45^{\circ}\cos30^{\circ}-\cos45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}\approx0.259\).
So \(AB=\frac{8}{0.259}\approx30.9\).

Step3: Find \(AC\)

Since \(\tan A=\frac{BC}{AC}\), then \(AC = \frac{BC}{\tan A}\).
We know that \(\tan15^{\circ}=\tan(45^{\circ}-30^{\circ})=\frac{\tan45^{\circ}-\tan30^{\circ}}{1 + \tan45^{\circ}\tan30^{\circ}}=\frac{1-\frac{\sqrt{3}}{3}}{1 + 1\times\frac{\sqrt{3}}{3}}=2-\sqrt{3}\approx0.268\).
Substitute \(BC = 8\) into the formula: \(AC=\frac{8}{0.268}\approx30.0\).

Answer:

\(AC = 30.0\), \(AB=30.9\)